Timeline for Is the matrix $\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}$ nonsingular?
Current License: CC BY-SA 3.0
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| Feb 12, 2017 at 1:06 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Jan 1, 2017 at 15:21 | comment | added | T. Amdeberhan | @RichardStanley: You may like to see my question, part 3, on SNF at mathoverflow.net/questions/258448/… | |
| Dec 30, 2016 at 22:13 | comment | added | T. Amdeberhan | The above proof appears simpler ... | |
| Dec 30, 2016 at 9:18 | comment | added | Christian Stump | I agree with @Suvrit that this is a special case of (3.13) in Krattenthaler's Advanced determinant calculus by setting $B=0, A=x, L_j = x-2j-\lambda$, and interchanging $i$ and $j$. Quote from the author's introduction In fact, I claim that about 80 % of the determinants that you meet in “real life,” and which can apparently be evaluated, are a special case of just the very first of these [lemmas]. | |
| Dec 30, 2016 at 3:16 | comment | added | Richard Stanley | @T.Amdeberhan: my conjecture seems also to be true for any positive integer $m$ and any integer $\lambda$. | |
| Dec 30, 2016 at 1:27 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Dec 29, 2016 at 23:05 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Dec 29, 2016 at 23:01 | comment | added | T. Amdeberhan | @RichardStanley: I've changed notation, so your conjecture is for $A_{2m-1}(x,0)$. Would it still hold for $A_n(x,\lambda)$? | |
| Dec 29, 2016 at 22:46 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
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| Dec 29, 2016 at 22:07 | comment | added | Richard Stanley | Conjecture. All the diagonal entries of the Smith normal of $A_m(x)$ over the ring $\mathbb{Q}[x]$ are squarefree (as polynomials in $x$). This uniquely determines the SNF. I checked this for $m\leq 10$ and could easily check some further cases. On the other hand, the eigenvalues do not look nice. The characteristic polynomial of $A_m(x)$ is irreducible for $1\leq m\leq 4$. | |
| Dec 29, 2016 at 16:41 | comment | added | Suvrit | It seems (3.13) in "advanced determinant calculus" contains this as a special case. | |
| Dec 29, 2016 at 15:46 | comment | added | Mark Wildon | Have you looked at the eigenvalues of $A_m(x)$? There may be a a generalization of Holte's result. | |
| Dec 29, 2016 at 8:26 | comment | added | Fedor Petrov | Often such identities are provable by induction using Desnanot–Jacobi (also known as Lewis Carroll's) identity en.wikipedia.org/wiki/Dodgson_condensation | |
| Dec 29, 2016 at 6:18 | history | answered | T. Amdeberhan | CC BY-SA 3.0 |